Question

The least value of $3x+4y$ for positive values of $x$ and $y$ subject to the conditions $x^2{y}^3=6$ is

• A. $4$
• B. $6$
• C. $8$
• D. $10$

Answer 1

The correct option is D

$10$

Find the least value of the given expression based on given conditions

It is given that $x^2{y}^3=6$, where $x>0,y>0$

And, the expression whose least value id to be calculated is $3x+4y$.

The above expression can be expanded as,

$\frac{3x}{2}+\frac{3x}{2}+\frac{4y}{3}+\frac{4y}{3}+\frac{4y}{3}$

Now, we know that the relation between the AM (Arithmetic Mean) and the GM (Geometric Mean) is,

$AM\ge GM$

Now, using the relation between the AM and the GM in the terms $\frac{3x}{2},\frac{3x}{2},\frac{4y}{3},\frac{4y}{3},\frac{4y}{3}$,

$\because$ $AM\ge GM$

$\Rightarrow$ $\left(\frac{\frac{3x}{2}+\frac{3x}{2}+\frac{4y}{3}+\frac{4y}{3}+\frac{4y}{3}}{5}\right)\ge {\left(\frac{3x}{2}\times \frac{3x}{2}\times \frac{4y}{3}\times \frac{4y}{3}\times \frac{4y}{3}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow$ $\left(\frac{\frac{3x+3x}{2}+\frac{4y+4y+4y}{3}}{5}\right)\ge {\left(\frac{3}{2}\times \frac{3}{2}\times \frac{4}{3}\times \frac{4}{3}\times \frac{4}{3}{x}^2{y}^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow$ $\left(\frac{3x+4y}{5}\right)\ge {\left(\frac{16}{3}{x}^2{y}^3\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow$ $\frac{3x+4y}{5}\ge {\left(\frac{16}{3}\times 6\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ …$\left[\because x^2{y}^3=6\right]$

$\Rightarrow$ $\frac{3x+4y}{5}\ge {\left(32\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow$ $\frac{3x+4y}{5}\ge {\left(2\times 2\times 2\times 2\times 2\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow$ $\frac{3x+4y}{5}\ge 2$

$\Rightarrow$ $3x+4y\ge 10$ … [multiplying by $5$ on both sides]

Hence, option (D) is the correct option.